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NEET Chemistry Chemical Kinetics / रासायनिक बलगतिकी Question Bank Self Evaluation Test - Chemical Kinetics

  • question_answer
    A catalyst lowers the activation energy of a certain reaction from 83.314 to \[75\text{ }kJ\text{ }mo{{l}^{-1}}\] at 500 K. What will be the rate of reaction as compared to uncatalysed reaction? Assume other things being equal.

    A) Double             

    B) 28 times

    C) 7.38 times         

    D) \[7.38\times {{10}^{3}}\]times

    Correct Answer: C

    Solution :

    [c] \[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{A{{e}^{-{{E}_{{{a}_{2}}}}/RT}}}{A{{e}^{-{{E}_{{{a}_{1}}}}/RT}}}={{e}^{({{E}_{{{a}_{1}}}}-{{E}_{{{a}_{2}}}})/RT}}\] \[2.303\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{{{a}_{1}}}}-{{E}_{a}}}{RT}\] \[=\frac{\left( 83.314-75 \right)\times {{10}^{3}}}{8.314\times 500}=2\] \[\log {{k}_{2}}=\frac{2}{2.303}=0.868\] Taking Antilog \[{{k}_{2}}=7.38\]


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